We have two problems to solve. (c) Simplify the expression $[5a^5b^2 \times 3(ab^3)^2] \div (15a^2b^8)$. (d) If $a$ and $b$ are whole numbers such that $a^b = 121$, evaluate $(a-1)^{b+1}$.

1. Problem Description

We have two problems to solve.

(c) Simplify the expression $[5a^5b^2 \times 3(ab^3)^2] \div (15a^2b^8)$.

(d) If $a$ and $b$ are whole numbers such that $a^b = 121$, evaluate $(a-1)^{b+1}$.

2. Solution Steps

(c) Simplify the expression $[5a^5b^2 \times 3(ab^3)^2] \div (15a^2b^8)$.

First, we simplify the term $(ab^3)^2$ by applying the power of a product rule, $(xy)^n = x^ny^n$.

$(ab^3)^2 = a^2(b^3)^2 = a^2b^6$.

Now substitute this back into the expression:

$[5a^5b^2 \times 3a^2b^6] \div (15a^2b^8)$.

Next, multiply the terms inside the brackets:

$5a^5b^2 \times 3a^2b^6 = (5 \times 3)(a^5 \times a^2)(b^2 \times b^6) = 15a^{5+2}b^{2+6} = 15a^7b^8$.

Now substitute this back into the expression:

$[15a^7b^8] \div (15a^2b^8)$.

Now, we divide:

$\frac{15a^7b^8}{15a^2b^8} = \frac{15}{15} \times \frac{a^7}{a^2} \times \frac{b^8}{b^8} = 1 \times a^{7-2} \times b^{8-8} = a^5b^0 = a^5 \times 1 = a^5$.

(d) If $a$ and $b$ are whole numbers such that $a^b = 121$, evaluate $(a-1)^{b+1}$.

We need to find whole numbers $a$ and $b$ such that $a^b = 121$.

We can express 121 as $11^2$. Thus $a = 11$ and $b = 2$.

We can also express 121 as $121^1$, then $a=121$ and $b=1$.

Another possibility is $a^b = (-11)^2 = 121$. However, since we are given that a and b are whole numbers, then $a = 11$ and $b = 2$, or $a = 121$ and $b = 1$.

Case 1: $a = 11$ and $b = 2$.

$(a-1)^{b+1} = (11-1)^{2+1} = (10)^3 = 1000$.

Case 2: $a = 121$ and $b = 1$.

$(a-1)^{b+1} = (121-1)^{1+1} = (120)^2 = 14400$.

Since the problem does not specify a unique solution for $a$ and $b$, we can assume that the question refers to the most obvious solution. $121 = 11^2$, thus $a = 11$ and $b = 2$.

$(a-1)^{b+1} = (11-1)^{2+1} = 10^3 = 1000$.

3. Final Answer

(c) $a^5$

(d) 1000